EM Field Strength in Curved Spacetime: Is it Unchanged?

[DuckAmuck]

TL;DR Summary

F_mu_nu does not change in the presence of a gravitational field?

It seems a gravitational field does not alter the electromagnetic field strength. Is this correct?
My reasoning:
With no gravity, field strength is:
$$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$$
Introduce gravity:
$$\partial_\mu A_\nu \rightarrow \nabla_\mu A_\nu = \partial_\mu A_\nu + \Gamma_{\mu\nu}^{\alpha} A_\alpha$$
Field strength then becomes:
$$F_{\mu\nu} = \nabla_\mu A_\nu - \nabla_\nu A_\mu$$
$$= \partial_\mu A_\nu - \partial_\nu A_\mu + \Gamma_{\mu\nu}^{\alpha} A_\alpha - \Gamma_{\nu\mu}^{\alpha} A_\alpha$$
$$= \partial_\mu A_\nu - \partial_\nu A_\mu$$
The Christoffel symbol terms cancel. So you end up with an unchanged field strength.
Is this correct? It seems to make sense as field strength is essentially the number of photons, and gravity should not change that.
Some further insight would be much appreciated.

 

[anuttarasammyak]

As for covariant component $F_{\mu\nu}$ your calculation is right.
However, this cancellation does not take place for $F^{\mu\nu}$ and $F^\mu_{\ \ \nu}$.

 

[pervect]

Summary:: F_mu_nu does not change in the presence of a gravitational field?

It seems a gravitational field does not alter the electromagnetic field strength. Is this correct?

I get

$$ \nabla_\mu A_\nu = \partial_\mu A_\nu - \Gamma^\alpha{\mu\nu} \, A_{\alpha}$$

But the sign difference doesn't matter to the symmetry, the symmetric Christoffel terms cancel out when you anti-symmetrize $\nabla_\mu A_\nu$ to get $F_{\mu\nu}$.

I came to similar, though less general, conclusiong regarding $F_{\mu\nu}$ in the Rindler metric.

But there are two cautions. As others have mentioned, this is only true for $F_{\mu\nu}$, not $F^{\mu\nu}$. And additionally, it's important to note that it's only true in a coordinate basis, which is the only basis where the expression you give for the covariant derivative is correct.

Using a non-coordinate basis is common, and even desirable. In particular, the readings of physical measuring instruments (such as the volts/meter reading of a co-located E field meter) will be given by the components of the Faraday tensor in a local orthonormal basis, which is not a coordinate basis if the space-time isn't flat Minkowskii.

See for instance https://www.physicsforums.com/threads/parallel-plate-capacitor-in-the-rindler-metric.994807/ for the Rindler case. To give the highlights, I computed that that the local voltage readings measured between the plates along a path of constant "height" (z-coordinate) depended on $g_{00}$, which varies with the "height" / z coordinate.

I also found that in a coordinate basis $F_{\mu\nu}$ has the same component values as it would in flat space-time, namely $F_{\mu\nu} = dt_\mu \wedge dz_\nu$ , where $\wedge$ is the "wedge product".

Note also that this was a result I derived, not a textbook result. However, while I did flail around a bit, I have confidence in my eventual conclusions.

 

[PeterDonis]

It seems a gravitational field does not alter the electromagnetic field strength.

That's not what your math is telling you. All that your math is telling you is that picking a curvilinear coordinate system does not change any physical observables. Which means you did the math correctly and got the expected right answer, since any change in coordinates should not affect any physical observables.

If you really want to "introduce gravity", you need to add curvature, not just nonzero connection coefficients. And then you have to deal with the fact that the EM field contains energy, and hence acts as a source of spacetime curvature. In other words, you need to look at self-consistent solutions of the Einstein-Maxwell Equations in which the spacetime geometry is something other than flat Minkowski spacetime, because the stress-energy of the EM field means the spacetime is no longer vacuum, and flat Minkowski spacetime is only a solution if the spacetime is vacuum (zero stress-energy).

Once you have done that, what you will find is that the question you are asking isn't really meaningful, because there is no way to compare "the electromagnetic field strength" in two different spacetime geometries to see if they are "the same" or not. For example, the solution for a "static electric charge" is Reissner-Nordstrom spacetime; but there is no way to meaningfully ask if "the presence of gravity" in Reissner-Nordstrom spacetime "changes the field" of a static charge, because there is no "without gravity" case to compare it to--the "without gravity" case would be flat Minkowski spacetime, but flat Minkowski spacetime with a static charge is not a self-consistent solution (since the static charge, as above, has nonzero stress-energy and Minkowski spacetime is not a valid spacetime geometry for nonzero stress-energy).

 

[anuttarasammyak]

And additionally, it's important to note that it's only true in a coordinate basis, which is the only basis where the expression you give for the covariant derivative is correct.

As you said does the familiar formula defining covariant derivative using $\Gamma^\sigma_{\mu\nu}$ holds only in some coordinates ? If so more than metric is given, what is necessary for this formula to hold?

 

[vanhees71]

Indeed, the connection between the four-potential and the Faraday tensor doesn't change, because alternating differential forms do not depend on the connection. You can define them on general differentiable manifolds without connection or (pseudo-)metric.

Nevertheless the remaining inhomogeneous Maxwell equations,
$$\nabla_{\mu} F^{\mu \nu}=\frac{1}{c} j^{\nu},$$
depend on the connection and/or the (pseudo-)metric.

That's also physically clear since one of the best known gravitational effects on em. waves is the bending of light, whose observational confirmation in 1919 through Eddington's solar-eclipse observations made Einstein to the first celebrity of the natural sciences. This implies that the gravitational interaction, described in GR by the curvature of spacetime, indeed affects the electromagnetic field. Another example is the gravitational red shift, which can be observed thanks to the Moessbauer effect.

 

[pervect]

As you said does the familiar formula defining covariant derivative using $\Gamma^\sigma_{\mu\nu}$ holds only in some coordinates ? If so more than metric is given, what is necessary for this formula to hold?

It's the choice of basis that matters in this case, not the coordinate choice.

An example might be helpful.

If we analyze a parallel plate capacitor in cartesian coordinates, we find that in units where c=1 and $2 \pi \epsilon_0 = 1$, we get

$$F_{\mu\nu} = \begin{bmatrix} 0 & \lambda & 0 & 0 \\ -\lambda & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

with the cartesian line element of $-dt^2 + dx^2 + dy^2 + dz^2$.

Now, we just argued that the line element doesn't matter, so let's change the line element to the line element in cylindrical coordinates rather than cartesian coordinates.

We do this by making a formal analogy t->t, x->r, y->$\theta$, z->z. The new line element becomes:

$$-dt^2 + dr^2 + r^2 d\theta^2 + dz^2$$

You could alternately use the line element

$$-dt^2 + dx^2 + x^2 dy^2 + dz^2$$.

This turns the parallel plate capacitor into a coaxial capacitor. We've just argued that the components of $F_{\mu\nu}$ are unchanged, i.e. that the components are$$F_{\mu\nu} = \begin{bmatrix} 0 & \lambda & 0 & 0 \\ -\lambda & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

But if we look up a typical undergraduate analysis of a coaxial capacitor, for instance http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capcyl.html

we see that by Gauss law that $E \propto \frac{\lambda}{r}$.

There are a few things here, as I write this, that I need to think over a bit, I think, as far as the resolution goes. But I think it's an interesting enough example that I'll just post this anyway.

 

[pervect]

I did a bit more on the problem I mentioned earlier, finding the radial electric field in cylindrical coordinates. I think it's a good exercise to go through to understand the limits of the remarks about the coordinate independence of the electromagnetic field.

The line element I used is the obvious one:

$$-dt^2 + dr^2 + r^2 d\theta^2 + dz^2$$

We assume a Faraday tensor of the form

$$F_{ab} = \begin{bmatrix} 0 & E(r) & 0 & 0 \\ -E(r) & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

in the coordinate co-basis dt, dr, d$\theta$, dz. It's the same in the orthonormal co-basis dt, dr, rd$\theta$ dz.

The equation $\nabla_a F^{ab}=0$ and/or the equation $\nabla^a F_{ab}=0$ give the equations E(r) must satisfy, but due to the raised index/indices, are NOT equivalent to $\partial_a F^{ab}=0$ or $\partial^a F_{ab}=0$.

Specifically, we find that
$$\frac{E(r)}{r} + \frac{\partial E(r)}{\partial r} = 0$$

The equations

$$\nabla_a F_{bc} + \nabla_b F_{ca} + \nabla_c F_{ab} = 0$$ are equivalent to the equations
$$\partial_a F_{bc} + \partial_b F_{ca} + \partial_c F_{ab} = 0$$

and are independent of the connection, but are satisfied for any function E(r), so they don't tell us what E(r) is.

If we introduce the Maxwell tensor, the Hodges dual of F, we can write

$$\nabla_a M_{bc} + \nabla_b M{ca} + \nabla_c M_{ab} = 0$$
and see that in the coordinate co-basis, $dt, dr, d\theta, dz$ the above equations are equivalent to
$$\partial_a M_{bc} + \partial_b M_{ca} + \partial_c M_{ab} = 0$$

We can set the above zero because there is no charge or current between the plates. And the lack of any raised indices is what makes the equation independent of the connection.

To be specific, we find that in the coordinate cobasis $dt, dr, d\theta, dz$

$$M_{ab} = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \lambda \\ 0 & 0 & -\lambda & 0 \end{bmatrix}$$

where $\lambda$ is some constant.

However, in the orthonormal co-basis $dt, dr, r\,d\theta, dz$, the components of M are not independent of r. Basically the nonzero components $\lambda d\theta dz$ in the coordinate basis become $\frac{\lambda}{r} (r d\theta) (dz)$ in the orthonormal cobasis. And the later is what gives us the physically measured electric field in volts/meter, not the components in the coordinate basis. So again, we recover the 1/r dependence of the electric field for the cylindrical capacitor.